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CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

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LEARNING OBJECTIVES Understand random variables Calculate means and variances Determine probabilities from cumulative distribution functions Understand the assumptions of the discrete probability distributions Calculate probabilities, determine means and variances for each of the discrete probability distributions

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Concept of Random Variable Summarize the outcome from a random experiment by a simple number Used to describe the possible outcomes Useful to associate a number with each outcome in the sample space Variable that associates a number with the outcome of a random experiment is referred to as a random variable

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Definition of Random Variables Function that assigns a real number to each outcome in the sample space Denoted by an uppercase letter such as X Value of the random variable is denoted by a lowercase letter such as x

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Two Types of Random Variables Discrete and Continuous Sometimes a measurement can assume any value in an interval of real numbers Said to be a continuous random variable –Examples: Electrical current, length, pressure, temperature, time, voltage, weight Sometimes the measurement is limited to integers Said to be a discrete random variable –Examples: No. of scratches on a surface, No. of calls received per day, and No. of transmitted bits received in error

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Probability Distribution Describes probabilities associated with the possible values of X Discrete Case –Specified by just a list of the possible values along with the probability of each Continuous Case –Used to describe the probability distribution Convenient to express the probability in terms of a formula

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Discrete Probability Distribution Definition –For a discrete random variable X with possible values x 1, x 2, x 3, …, x n, the probability mass function (PMF) is f(x i ) = P(X = x i ) f(x i ) 0 for all x i

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Discrete Probability Distribution Examples To better understand the PMF, consider Example 1 –Tossing a 6-sided die –f(x)=1/6 for X=1,2,3,4,5,6 Example 2 –Check whether the following can serve as probability distribution –1. f(x)= (x-2)/2 for x=1,2,3,4; –2. h(x)= x 2 /25 for x=0,1,2,3,4;

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Class Problem The sample space of a random experiment is {a,b,c,d,e,f}, and each outcome is equally likely. A random variable is defined as follows: Determine the probability mass function of X a)f(0)=P(X=0)=1/6+1/6=1/3 b)f(1.5)=P(X=1.5)=1/6+1/6=1/3 c)f(3)=P(X=3)=1/6 Outcomeabcdef x001.5 23

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Cumulative Distribution Function Useful to provide cumulative probabilities such as P(X x) Cumulative distribution function (CDF) of a discrete random variable X, denoted as F(x), is F(x) has the following properties: –F(x)= P(X x)= – 0 F(x) 1 –If x y, then F(x) F(y)

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Cumulative Distribution Function- Example Determine the cumulative distribution of the random variable for which each outcome is equally likely in the following sample space: Outcomeab c def x001.5 23

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Solution Outcomeab c def x001.5 23 f(x)1/6

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Calculating Probabilities from PMF and CDF Interested to determine –Probabilities from cumulative distribution functions –Cumulative distribution functions from probability mass functions –Vise versa Consider the following example

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Example Consider the following PMF Determine the CDF y1234 f(y).4.3.2.1 y f(y) 0.4 1 F(y) y 1

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Example Consider the following CDF Note that Y can take on 1, 2, 3, or 4. Determine the PMF –P(Y=1)=0.4-0.0=0.4 –P(Y=2)=0.7-0.4=0.3, P(Y=3)=0.9-0.7=0.2, P(Y=4)=1.0-0.9=0.1 –Hence, y1234 f(y).4.3.2.1

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Mean of a Discrete Random Variable PMF provides complete information about the properties of a random variable Useful to have some summary measures of these properties Mean or expectation –Denoted by E(X) or represents an average value of the random variable If a random variable takes on the values x 1, x 2,…, or x k with the probabilities f(x 1 ), f(x 2 ), …., and f(x k ), its mean is = E(X)= x 1.f(x 1 ) + x 2.f(x 2 )+...+ x k.f(x k ) –Or = E(X)=

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Examples Find the mean of the probability distribution of the number of heads obtained in three flips of a coin –Probabilities 1/8, 3/8, 3/8, and 1/8 = (0)1/8 + (1)3/8 + (2)3/8 + (3) 1/8 = 3/2 Suppose Y is the total showing on a pair of dice, find the mean of the probability distribution –Calculated as

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Variance of a Discrete Random Variable Another important measure of the distribution of a random variable Measures the spread or variability Whereas mean measures central, the variance measures the deviation Denoted as 2 or V(x) 2 = V(X)=E(X- ) 2 = Standard deviation of X is =[V(X)] 1/2

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Examples Find the variance of the probability distribution of the number of heads obtained in three flips of a coin –Variance 2 = (0-3/2) 2 (1/8)+(1-3/2) 2 (3/8)+(2-3/2) 2 (3/8)+(3- 3/2) 2 (1/8)=0.75 Find the variance of the total showing on a pair of dice –Variance 2 = V(X)= (2-7) 2( 1/36) + (3-7) 2 (2/36)+ … + (12-7) 2 (1/36) =

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Class Problem If the range of X is the set {0, 1, 2, 3, 4} and P(X=x)= 0.2, determine the mean and variance of the random variable Solution –X can take values of 0,1, 2, 3 or 4 –Summation of x f(x) –Hence, mean=2 –Variance can be calculated as –V(X)= (0-2) 2 (0.2)+(1-2) 2 (0.2)+…=

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Uniform Discrete Distribution If X assumes the values x 1,x 2,….,x n with equal probability, then it has a discrete uniform distribution Probability mass function f(x i )= 1/n for x i = x 1,x 2,….,x n Mean E[X]= Variance V(X)=

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Examples When a light bulb is selected randomly from a box containing a red, a blue, a white, and a yellow bulb –Sample space is {red, blue, white, yellow} with probability 1/4 –Hence, f(x)=1/4 When a die is tossed –Sample space is {1,2,3,4,5,6} with probability 1/6 –f(x)=1/6 –E[X]= (1*1/6+ 2*1/6+ 3*1/6+ 4*1/6+ 5*1/6+6*1/6)=3.5 –V[X]=(1-3.5) 2 /6 + (2-3.5) 2 /6 + …+ (6-3.5) 2 /6 =35/12

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Binomial Distribution Two possible outcomes labeled success or failure Referred to as a Bernoulli process Conditions –Consists of n repeated trails –Results in two possible outcomes (Success or Failure) –Probability of success, denoted by p, remains constant –Trials are independent

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Examples Tossing a coin 10 times 1.There are 10 trials, and they are identical 2.Each trial has only two outcomes 3.Probability of getting a H is 0.5 in each trial 4.Trials (tosses) are independent In an operation, 5% of all machined parts are defective. 3 parts are randomly selected from the production line to determine if each of them is defective or good 1.Three identical trials 2.Each trial has two outcomes 3.p=0.05 4.Independent trials

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PMF of the Binomial Distribution Binomial distribution with parameters p and n=1,2,… PMF –f(x)= p x (1-p) n-x –Where x =0,1, 2,…,n. – E[X]= np V (X)=np(1-p)=npq

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Example In an operation, 5% of all parts machined by a firm are defective If three parts are randomly selected from the production line, what is the probability that exactly one of them will be defective? Solution –n=3, x=1, and p=0.05 –Substituting in the PMF P(X=1)= f(1) = 0.05 1 (1-0.05) 3-1 P(X=1)=3*0.05 1 (1-0.05) 3-1 = 0.1354

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Class Problem The random variable X has a binomial distribution with n=10 and p=0.01, determine the following probabilities P(X=5) = (0.01) 5 (1-0.01) 10-5) = P(X 2) =P(X=0)+P(X=1)+P(X=2)=0.904 +0.091+0.00045 =0.99 P(3 X<5) =P(X=3)+ P(X=4)=1.14

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Geometric Distribution Closely related to the binomial experiment Trials are conducted until a success is obtained Let X denote the number of trials PMF –P(X = x) = f(x) = (1-p) x-1 p x = 1,2,3,… Mean and Variance – =E(x)=1/p and 2 =V(x)=(1-p)/p 2

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Example The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8 Assume the trials are independent –What is the probability the first successful alignment requires exactly four trials? –What is the probability that the first successful alignment requires at most four trials? –What is the probability that the first successful alignment requires at least four trials?

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Example Let X denote the number of trials to obtain in the first successful alignment. Then X is a geometric random variable with p = 0.8 Solution P(X = 4) = f(4) = 0.2 3 (0.8) = 0.0064 P(X 4) = P(X=1) + P(X = 2) + P(X =3) + P(X = 4)=0.9984 P(X 4) = 1 P(X < 4) = 1 0.992 = 0.008

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Class Problem Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal (calls are independent) –What is the probability that your first call that connects is your tenth call? –What is the probability that it requires more than five calls for you to connect? –What is the mean number of calls needed to connect?

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Solutions Let X denote the number of calls needed to obtain a connection Then, X is a geometric random variable with p=0.02 P(X = 10) = P(X>5) = E(X) =1/0.02 = 50

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Negative Binomial Distribution Based on an experiment –Consists of a sequence of independent trials –Result in either a “S” or “F” –Probability of success is constant from trial to trial –Continues until a total of r success have been observed PMF P(X = x) = f(x) = (1-p) x-r p r Mean and Variance E[X] = r/p and V (X) = r(1-p)/p 2

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Example Suppose that X is a negative binomial random variable with p=0.2 and r=4. Determine the following E(X), P(X = 20), P(X = 19), P(X = 21) Solution E(X) = r/p =4/0.2 = 20 P(X = 20) = (1-p) x-r p r = P(X = 19) = P(X = 21) =

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Hypergeometric Distribution Assumptions –Population consists of N objects (finite) –Each classified as a “S” or “F” and K success –n individuals is selected without replacement Random variable of interest is X, the number of successes in the sample

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PMF of the Hypergeometric Distribution X=number of successes in a sample of size n drawn from a population consisting of K successes and N-K failures PMF is given –for x satisfying max (0, n-N+K) x min (n,K) Mean and Variance E[X] = n and V (X) = n

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Example A shipment of 20 machined parts contains 5 that are defective If 10 of them are randomly chosen for inspection, what is the probability that 2 of the 10 will be defective? Solution x = 2, n = 10, k = 5, and N=20 f(x=2)= = 0.348

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Class Problem A lot of 75 washers contains 5 in which the variability in thickness around the circumference of the washer is unacceptable A sample of 10 washers is selected at random, without replacement. –What is the probability that none of the unacceptable washers is in the sample? –What is the probability that at least one unacceptable washer is in the sample? –What is the probability that exactly one unacceptable washer is in the sample? –What is the mean number of unacceptable washers in the sample?

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Solution Let X denote the number of unacceptable washers in the sample of 10 P(X = 0) P(X 1) =1- P(X=0)=0.5214 P(X = 1)= E(X) =10*5/75=2/3

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Poisson Distribution Binomial, hypergeometric, and negative binomial distributions start with an experiment consisting of trials Based on the number of outcomes occurring during a given time interval or in a specified regions Examples –# of accidents that occur on a given highway during a 1- week period –# of customers coming to a bank during a 1-hour interval –# of TVs sold at a department store during a given week –# of breakdowns of a washing machine per month

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Conditions Consider the # of breakdowns of a washing machine per month example –Each breakdown is called an occurrence –Occurrences are random that they do not follow any pattern (unpredictable) –Occurrence is always considered with respect to an interval (one month)

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The Probability Mass Distribution X = number of counts in the interval Poisson random variable with > 0 PMF f(x)= x=0,1,2, Mean and Variance E[X] =, V (X) =

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Example If a bank gets on average = 6 bad checks per day, what are the probabilities that it will receive four bad checks on any given day?10 bad checks on any two consecutive days? Solution x = 4 and = 6, then f(4) = = 0.135 = 12 and x = 10, then f(10) = = 0.105

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Class Problem The number of failures of a testing instrument from contamination particle on the product is a Poisson random variable with a mean of 0.02 failure per hour. –What is the probability that the instrument does not fail in an 8-hour shift? –What is the probability of at least one failure in one 24- hour day?

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Solution a)Let X denote the failure in 8 hours. Then, X has a Poisson distribution with =0.16 P(X=0)=0.8521 b)Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with =0.48 P(Y 1) = 1-P(Y = 0) =0.3812

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The Poisson Approximation To The Binomial Distribution When n is large and p is small, binomial probabilities are often approximates by f(x)= for x = 0,1,2, … with = np

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Example Assume 5% of the books at a certain bindery have defective bindings. Find the probability that 2 of 100 books bound by this bindery will have defective bindings, using –The binomial distribution –The Poisson distribution

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Solution Using Binomial x = 2, n = 100, and p = 0.05 f(x) = (0.05) 2 (0.95) 98 = 0.081 Using Poisson x = 2, = np = 100*0.05 = 5 f(2) = = 0.084

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Next agenda Continue our development of probability distributions with a discussion of several important continuous distributions

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